# Batch Sparging – Making sense of the numbers

This weekend I took the time to take extensive
extract and volume measurements during a 2 sparge batch sparging
process here is the data and an analysis of that data:

• grist weight 5.6 kg
• total laboratory extract of that grist is  80% of 5.6 kg -> 4.5 kg
• water added to mash: 15.5 l (cold)
• extract of the first running in the kettle 22.5% (% extract is equal to Plato)
• volume of the first runnings in the kettle 9.75l at 65C -> 9.6l (cold)
• extract of the 2nd runnings: 11.75%
• volume in kettle after 2nd running: 20l at 75C -> 19.6l (cold)
• extract of the 3rd runnings: 7.4%
• volume in kettle after 3rd running (pre-boil volume): 26l at 90C -> 25l (cold)
• extract in kettle after 3rd running (pre-boil extract): 14.6%

The first analysis was for the extraction efficiency of the mash. The definition of extract percentages is:

(1)  E = 100% * m_extract / ( m_water + m_extract)

If
we want to know how much extract exist in a given wort of known extract
content that has been created with a known  amount of water we can do
this by rearanging (1) to

(2) m_extract = (m_water * E / 100%) / (1 – E / 100%)

(3) m_extract = (15.5kg * 0.225) / (1 – 0.225) = 4.5 kg

This
means that all of the extract available in the grain has been extracted
in the mash (100% extraction efficiency). This was confirmed by a
negative iodine test of the wort and the spent grain. I.e. no native
starch was left.

Since batch sparging was
used, a simple model can be used to calculate the lauter efficiency.
lauter efficiency * extraction efficiency is the brewhouse efficiency.
For that model we need the amount of wort that is held back in the
lauter tun after each run-off. But this is not simply the amount of
water used for the mash minus the amount of first wort collected
because the volume of the wort increases when the extract is dissolved.
To get that volume we can use this formula which is the weight of
extract dissolved in a given volume of known gravity wort:

(4) m_extract = ( E / 100% ) * SG * V_wort

SG is the specific gravity and it will be estimated with 1+E*0.004.Rearranged to V_wort we get

(5) V_wort = m_extract /  ((E/100%) * SG)

(6) V_wort = 4.5 kg / (0.225 * 1.090) = 18.3 l

This
means the 15.5 l water and 4.5 kg extract from the 5.6 kg grain made
18.3 l of 22.5% wort. 9.6l of that wort were collected after the first
run-off which indicates that 8.7 l are held back in the mash.

Batch
sparing is a process of successive dilution of the wort remaining in
the grain and running it off. This can be modeled mathematically and
has bee analyzed here. But since not all run-offs were of equal size, lets just calculate the efficiency step by step:

The first run-off will extract this percentage of the extract from the mash:

(7) Eff_1st = v_1st_runoff / (v_1st_runoff + v_wort_in_grain)

(8) Eff_1st = 9.6l / (9.6l + 8.7l) = 0.52 = 52 %

If
52% were recovered by the 1st run-off, then 48% of the extract are
still in the lauter tun. This extract is dilluted by the sparge water
and run off. The volume of the 2nd run_off is 19.6l – 9.6l = 10l and
the efficiency of that run-off is:

(9) Eff_2nd = v_2nd_run_off / (v_2nd_run_off + v_wort_in_grain)

(10) Eff_2nd = 10l / (10l + 8.7l) =0.53 = 53%

Using
this and the fact that the 2nd run-off was only able to draw from 48%
of the extract we can determine the combined efficiency from the 1st
and 2nd run off as:

(11) Eff_1st_and_2nd = 52% + 48% * 53% = 78 %

78%
of the extract are now in the boil kettle. This leaves 22% in the
lauter tun. With a 3rd run off size of 5.4 l we find the efficiency of
that run-off as

(12) Eff_3rd = 5.4 / (5.4 + 8.7) = 0.38 = 38%

and the combined efficiency of all 3 run-offs as:

(13) Eff_1st_2nd_3rd =  52% + 48% * 53% + 22% * 38% = 0.86 = 86%

This
means that with the given run-off sizes, number of sparges and amount
of wort left in the grain, an a lauter efficiency of 86% is to be
expected.

The actual efficiency into the boiler is the following:

(14) Eff_kettle = V_kettle * E * SG / (m_grain * 0.8)

the 0.8 represents the 80% laboratory extract of the grain.

(15) Eff_kettle = 25l * 0.146 * 1.058 l/kg / (5.6 kg * 0.8) = 86%

Since
the Efficiency is the product of extraction efficiency and lauter
efficiency and the extraction efficiency was determined to be 100%, the
actual lauter efficiency must have been 86%, which matches the
theoretical result very well. As a result no efficiency was lost due to
process inefficiencies and to increase that efficiency the following
process parameters could be changed:

• more sparge water: this would lead to a larger pre boil volume and longer or stonger boils and may not be desired
• less wort kept in the grain: This mash was done with conditioned
malt which makes for a"fluffier" mash. Such a mash may hold more wort
and I wonder if an unconditioned mash may result in less wort being
held back and thus increasing the efficiency
• equalize the run-offs: the boost expected from that is very low. Se here.
• fly
sparging: this method follows a different principle and should yield
better efficiencies when done properly. But in addition to more time,
it also needs a better lautertun which I don't have.

So, 86% for that beer is fine with me.